SATHEE: Rotation 2 Question 24

Rotation 2 Question 24

24. A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_{0}$ , at an angle of $45^{\circ}$ to the horizontal. Find the magnitude and direction of the angular momentum of the particle about $P$ at time $t = \frac{v_{0}}{g}$ .

$(1984, 6 M)$

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Answer:

Correct Answer: 24. $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ in a direction perpendicular to paper inwards

Solution:

In terms of $\hat{i}, \hat{j}$ and $\hat{k}$

$\begin{aligned} u = \frac{v_{0}}{\sqrt{2}} \hat{i} + \frac{v_{0}}{\sqrt{2}} \hat{j} \\ \Rightarrow a = - g \hat{j} \\ t = \frac{v_{0}}{g} \\ = \frac{v_{0}^{2}}{\sqrt{2} g} \hat{i} + \frac{v_{0}^{2}}{\sqrt{2} g} \hat{j} - \frac{v_{0}^{2}}{2 g} \hat{j} \end{aligned}$

Now, angular momentum about point $P$ at given time

$\begin{aligned} L & = m (r \times v) \\ = m - \frac{v_{0}^{3}}{\sqrt{2} g} + \frac{v_{0}^{3}}{2 g} - \frac{v_{0}^{3}}{2 g} + \frac{v_{0}^{3}}{2 \sqrt{2} g} \hat{k} \\ = - \frac{m v_{0}^{3}}{2 \sqrt{2} g} \hat{k} \end{aligned}$

Thus, magnitude of angular momentum is $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ in $- \hat{k}$ direction i.e. in a direction perpendicular to paper inwards

Rotation 2 Question 24

24. A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_{0}$ , at an angle of $45^{\circ}$ to the horizontal. Find the magnitude and direction of the angular momentum of the particle about $P$ at time $t = \frac{v_{0}}{g}$ .

Answer:

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Rotation 2 Question 24

24. A particle is projected at time t=0 from a point P on the ground with a speed v0, at an angle of 45∘ to the horizontal. Find the magnitude and direction of the angular momentum of the particle about P at time t=v0g.

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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24. A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_{0}$ , at an angle of $45^{\circ}$ to the horizontal. Find the magnitude and direction of the angular momentum of the particle about $P$ at time $t = \frac{v_{0}}{g}$ .